how to do big x

We’ve found a quicker way to multiply really big numbers

how to do big x

Associate professor in mathematics, UNSW Sydney

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Multiplication of two numbers is easy, right?

At primary school we learn how to do long multiplication like this:

how to do big x

Methods similar to this go back thousands of years, at least to the ancient Sumerians and Egyptians.

But is this really the best way to multiply two big numbers together?

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In long multiplication, we have to multiply every digit of the first number by every digit of the second number. If the two numbers each have N digits, that’s N 2 (or N x N ) multiplications altogether. In the example above, N is 3, and we had to do 3 2 = 9 multiplications.

Around 1956, the famous Soviet mathematician Andrey Kolmogorov conjectured that this is the best possible way to multiply two numbers together.

In other words, no matter how you arrange your calculations, the amount of work you have to do will be proportional to at least N 2 . Twice as many digits means four times as much work.

Kolmogorov felt that if a short cut was possible, surely it would have already been discovered. After all, people have been multiplying numbers for thousands of years.

This is a superb example of the logical fallacy known as “the argument from ignorance”.

A quicker way

Just a few years later, Kolmogorov’s conjecture was shown to be spectacularly wrong.

In 1960, Anatoly Karatsuba, a 23-year-old mathematics student in Russia, discovered a sneaky algebraic trick that reduces the number of multiplications needed.

For example, to multiply four-digit numbers, instead of needing 4 2 = 16 multiplications, Karatsuba’s method gets away with only nine. When using his method, twice as many digits means only three times as much work.

This stacks up to an impressive advantage as the numbers get bigger. For numbers with a thousand digits, Karatsuba’s method needs about 17 times fewer multiplications than long multiplication.

But why on earth would anyone want to multiply such big numbers together?

In fact, there are a tremendous number of applications. One of the most visible and economically significant is in cryptography .

Big numbers in real life

Every time you engage in encrypted communication on the internet — for example, access your banking website or perform a web search — your device performs a head-spinning number of multiplications, involving numbers with hundreds or even thousands of digits.

Very likely your device uses Karatsuba’s trick for this arithmetic. This is all part of the amazing software ecosystem that keeps our web pages loading as snappily as possible.

For some more esoteric applications, mathematicians have to deal with even larger numbers, with millions, billions or even trillions of digits. For such enormous numbers, even Karatsuba’s algorithm is too slow.

A real breakthrough came in 1971 with the work of the German mathematicians Arnold Schönhage and Volker Strassen. They explained how to use the recently published fast Fourier transform ( FFT ) to multiply huge numbers efficiently. Their method is routinely used by mathematicians today to handle numbers in the billions of digits.

The FFT is one of the most important algorithms of the 20th century. One application familiar in daily life is digital audio: whenever you listen to MP3s, music streaming services or digital radio, FFTs handle the audio decoding behind the scenes.

An even quicker way?

In their 1971 paper , Schönhage and Strassen also made a striking conjecture. To explain, I’ll have to get a bit technical for a moment.

The first half of their conjecture is that it should be possible to multiply N -digit numbers using a number of basic operations that is proportional to at most N log ( N ) (that’s N times the natural logarithm of N ).

Their own algorithm did not quite reach this target; they were too slow by a factor of log (log N ) (the logarithm of the logarithm of N ). Nevertheless, their intuition led them to suspect that they were missing something, and that N log ( N ) should be feasible.

In the decades since 1971, a few researchers have found improvements to Schönhage and Strassen’s algorithm. Notably, an algorithm designed by Martin Fürer in 2007 came agonisingly close to the elusive N log ( N ).

The second (and much more difficult) part of their conjecture is that N log ( N ) should be the fundamental speed limit — that no possible multiplication algorithm could do better than this.

Sound familiar?

Have we reached the limit?

A few weeks ago, Joris van der Hoeven and I posted a research paper describing a new multiplication algorithm that finally reaches the N log ( N ) holy grail, thus settling the “easy” part of the Schönhage–Strassen conjecture.

The paper has not yet been peer-reviewed, so some caution is warranted. It is standard practice in mathematics to disseminate research results before they have undergone peer review.

Instead of using one-dimensional FFTs — the staple of all work on this problem since 1971 — our algorithm relies on multidimensional FFTs. These gadgets are nothing new: the widely-used JPEG image format depends on 2-dimensional FFTs, and 3-dimensional FFTs have many applications in physics and engineering.

In our paper, we use FFTs with 1,729 dimensions. This is tricky to visualise, but mathematically no more troublesome than the 2-dimensional case.

Really, really big numbers

The new algorithm is not really practical in its current form, because the proof given in our paper only works for ludicrously large numbers. Even if each digit was written on a hydrogen atom, there would not be nearly enough room available in the observable universe to write them down.

Read more: Why do we need to know about prime numbers with millions of digits?

On the other hand, we are hopeful that with further refinements, the algorithm might become practical for numbers with merely billions or trillions of digits. If so, it may well become an indispensable tool in the computational mathematician’s arsenal.

If the full Schönhage–Strassen conjecture is correct, then from a theoretical point of view, the new algorithm is the end of the road – it is not possible to do any better.

Personally, I would be very surprised if the conjecture turned out to be wrong. But we shouldn’t forget what happened to Kolmogorov. Mathematics can sometimes throw up surprises.

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Factoring Polynomials With Large Coefficients: Factoring by Extraction

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  • Aditya Virani
  • Dr. Bela Palnitkar

Factoring polynomials, in general, is quite difficult, but some special ones can be factored using certain tricks.

Factorizing Quadratics with Large Numbers

Factorizing high-degree polynomials with large numbers.

Today, I will discuss how to factor polynomials with large coefficients such as \(3x^2+10x-1000\) with ease. I know that this will be a long note, but I feel that it is worth reading everything including the generalized form at the bottom except for the proof (unless you want to).

While sitting in my math class today, I discovered a trick to factoring second-degree polynomials with large or irrational second and third coefficients. For example, try factoring \(3x^2+10x-1000\). It's relatively simple to factor it to \((3x-50)(x+20),\) but that would take a little while or at least longer than the way that I'm about to discuss.

We begin with the expression \(3x^2+10x-1000\). Then we divide the second coefficient by 10 and the third by 100, and we are left with the expression \(3x^2+x-10,\) which we can easily factor to \((3x-5)(x+2)\). Finally, we multiply the second term in each factor by 10 and have \((3x-50)(x+20)\). Looks familiar, doesn't it?

Basically, what I have done is that I divided the second coefficient by any one of its factors (in this case 10) and then divided the third coefficient by the square of that factor while leaving the first untouched.

This method applies to irrational and imaginary coefficients as well:

Factorize \(4x^2+8\sqrt2x+8\). Factor out \(2\sqrt2\) from the second coefficient and 8 from the third, and then we are left with \(4x^2+4x+1=(2x+1)^2.\) By multiplying back \(2\sqrt2\) to the second term in each factor, which in this case is \((2x+1),\) we have \[\left(2x+2\sqrt2\right)^2. \ _\square\]

This also works in reverse from the first coefficient:

Factorize \(25x^2-60x+36\). We can factor out 5 from the middle coefficient and 25 from the first, and then we are left with \(x^2-12x+36\). Next, we can factor out 6 from the second coefficient and 36 from the final coefficient, being left with \(x^2-2x+1=(x-1)^2\). Finally, we factor back in 5 to the first coefficient and 6 to the last one to obtain \((5x-6)^2\). \(_\square\)

This also works for unfactorable expressions. This method will not make unfactorable equations factorable; however, it will make the quadratic formula much easier to use. This is a little tougher to do because, depending on which way you factor a number out, the formula changes.

Case 1: \(ax^2+bx+c\Rightarrow ax^2+\frac{bx}{d}+\frac{c}{d^2}\).

This changes the quadratic equation to

\[\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{2a}=\dfrac{-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}}{2a}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a\color{blue}{\textbf{d}}}.\]

Thus, once our answer is achieved, we must multiply the answer by the number \(\color{blue}{\textbf{d}}\) that we extracted at the start.

Factorize \(x^2+60x+2025\). By factoring out \(15\) from the second coefficient and \(15^2=225\) from the final coefficient, we have \[\begin{align} x^2+\frac{60}{15}x+\frac{2025}{225} &=x^2+4x+9 \\ &=\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\\ &=\left(x-\big(-2+\sqrt{5}i\big)\right)\left(x-\big(-2-\sqrt{5}i\big)\right). \end{align}\] Now, by multiplying back \(15,\) we have \[\begin{align} x^2+60x+2025 &=\left(x-\big(-(15)2+(15)\sqrt{5}i\big)\right)\left(x-\big(-(15)2-(15)\sqrt{5}i\big)\right)\\ &=\left(x-\big(-30+15\sqrt{5}i\big)\right)\left(x-\big(-30-15\sqrt{5}i\big)\right).\ _\square \end{align}\]

Case 2: \(ax^2+bx+c\Rightarrow \frac{ax^2}{d^2}+\frac{bx}{d}+c\).

This changes the quadratic to

\[\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{\frac{2a}{d^2}}=\dfrac{d^2\left(-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}\right)}{2a}=\frac{{\color{blue}{\textbf{d}}}\big(-b\pm \sqrt{b^2-4ac}\big)}{2a}.\]

Thus, once our answer is achieved, we must divide the answer by the number \(\color{blue}{\textbf{d}}\) that we extracted at the start.

Now, we shall prove why this works:

Make the general expression \(ax^2+bx+c,\) which can be factored into \((dx+e)(fx+g).\) This means that \(a=df, b=dg+ef,\) and \(c=eg.\) The last step of our method requires us to multiply both of the second coefficients in our binomials by \(n\) \((n\) being the number that we factored out of \(b).\) So our expression becomes \(\big(dx+\frac{e}{n}\big)\big(fx+\frac{g}{n}\big),\) which means that \(a=df, \frac{b}{n}=\frac{dg+ef}{n}, \frac{c}{n^2}=\frac{eg}{n^2}\). This is why we factor out \(n\) and \(n^2\) from the second and third coefficients, respectively. \(_\square\)

This rule can actually apply to polynomials of any degree. Unfortunately, the higher the degree of the polynomial, the less convenient this becomes.

But, say, we have a polynomial with degree \(n\), which can be factored into \((a_1x+k_1)(a_2x+k_2)(a_3x+k_3)\cdots (a_nx+k_n)\). This does not necessarily have integer or real coefficients. Next, we assume that all coefficients after the second (inclusive) are divisible by \(r^t,\) where \(t\) represents the location of \(r\) relative to the second coefficient.

This means that we have \(\displaystyle\prod_{t=0}^n (a_tx+b_tr),\) which is equivalent to \(\displaystyle\sum_{t=0}^n \left(r^{n-t}\big(p_tx^t\big)\right).\) Upon expansion, we get

\[p_nx^n+r\big(p_{n-1}x^{n-1}\big)+r^2\big(p_{n-2}x^{n-2}\big)+r^3\big(p_{n-3}x^{n-3}\big)+\cdots+r^{n-1}\big(p_1x^1\big)+r^n(p_0).\]

As we can see, this method works whenever every coefficient after the first is divisible by \(r^t.\)

A useful method for seeing if this works is by prime-factoring every coefficient after the second and looking for a common term. That has multiple powers throughout. Say we have \(x^4-3x^3-63x^2+27x+486\). First, we prime-factor the numbers to get \(x^4-3x^3-3^2\cdot7x^2+3^3x+2\cdot3^5.\) As we can see, they share 3 in increasing powers. Therefore, we can eliminate 3 in increasing powers from each coefficient and are left with

\[\begin{align} x^4-x^3-7x^2+x+6&=x^3(x-1)-(7x+6)(x-1)\\ &=\left(x\big(x^2-1\big)-6(x+1)\right)(x-1)\\ &= \big(x^2-x-6\big)(x+1)(x-1)\\ &=(x-3)(x+2)(x+1)(x-1). \end{align}\]

Finally, we factor back in the three that we took out at the start and are left with \((x-9)(x+6)(x+3)(x-3).\)

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How to Solve Large Exponents

How to Solve Large Exponents

How to Factor Trinomials With the Diamond Method

As with most problems in basic algebra, solving large exponents requires factoring. If you factor the exponent down until all the factors are prime numbers – a process called prime factorization – you can then apply the power rule of exponents to solve the problem. Additionally, you can break the exponent down by addition rather than multiplication and apply the product rule for exponents to solve the problem. A little practice will help you predict which method will be easiest for the problem you are faced with.

Find the prime factors of the exponent. Example: 6 24

Use the power rule for exponents to set up the problem. The power rule states:

Solve the problem from the inside out.

Break the exponent down into a sum. Make sure the components are small enough to work with as exponents and do not include 1 or 0.

Example: 6 24

Use the product rule of exponents to set up the problem. The product rule states:

Solve the problem.

Things You'll Need

For some problems, a combination of both techniques may make the problem easier. For example: ​ x ​ 21 = (​ x ​ 7 ) 3 (power rule), and ​ x ​ 7 = ​ x ​ 3 × ​ x ​ 2 × ​ x ​ 2 (product rule). Combining the two, you get: ​ x ​ 21 = (​ x ​ 3 × ​ x ​ 2 × ​ x ​ 2 ) 3

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  • West Texas A & M University
  • Mesa Community College: A Review of the Rules for Exponents
  • For some problems, a combination of both techniques may make the problem easier. For example: x^21 = (x^7)^3 (power rule), and x^7 = x^3 * x^2 * x^2 (product rule). Combining the two, you get: x^21 = (x^3 * x^2 * x^2)^3

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How to factor faster with the X method?

  • Thread starter Guest
  • Start date Oct 31, 2006
  • Oct 31, 2006

stapel

Super Moderator

I've never heard of the "X method", and it doesn't seem to be particularly helpful. When dealing with big numbers, just grab your calculator and see what you can get. You multiplied 25 = 5×5 and 49 = 7×7, so you know that the only divisors of 25×49 = 1225 are going to be multiples of 5 and/or 7. On the off chance that this is a perfect square (since 25x<sup>2</sup> = (5x)<sup>2</sup> and 49 = 7<sup>2</sup>), check that first: . . . . . (2)(5x)(7) = 70x There ya go: it's a perfect-square trinomial, and you're done: . . . . . (5x + 7)<sup>2</sup> But, in general, just plug-n-chug, checking for (in this case) sums: . . . . . 1, 1225: sum of 1226 - NO . . . . . 5, 245: sum of 250 - NO . . . . . 7, 175: sum of 182 - NO . . . . . 25, 49: sum of 74 - NO . . . . . 35, 35: sum of 70 - YES Then factor as usual. Eliz.  

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  • Nov 1, 2006

Soroban and Stapel, thanks a ton!  

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7.3: The Shell Method

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Often a given problem can be solved in more than one way. A particular method may be chosen out of convenience, personal preference, or perhaps necessity. Ultimately, it is good to have options.

The previous section introduced the Disk and Washer Methods, which computed the volume of solids of revolution by integrating the cross--sectional area of the solid. This section develops another method of computing volume, the Shell Method . Instead of slicing the solid perpendicular to the axis of rotation creating cross-sections, we now slice it parallel to the axis of rotation, creating "shells."

Consider Figure \(\PageIndex{1}\), where the region shown in (a) is rotated around the \(y\)-axis forming the solid shown in (b). A small slice of the region is drawn in (a), parallel to the axis of rotation. When the region is rotated, this thin slice forms a cylindrical shell , as pictured in part (c) of the figure. The previous section approximated a solid with lots of thin disks (or washers); we now approximate a solid with many thin cylindrical shells.

7.3.1a.png

Figure \(\PageIndex{1}\): Introducing the Shell Method.

Figure \(\PageIndex{1}\) (d): A dynamic version of this figure created using CalcPlot3D.

To compute the volume of one shell, first consider the paper label on a soup can with radius \(r\) and height \(h\). What is the area of this label? A simple way of determining this is to cut the label and lay it out flat, forming a rectangle with height \(h\) and length \(2\pi r\). Thus the area is \(A = 2\pi rh\); see Figure \(\PageIndex{2a}\).

Do a similar process with a cylindrical shell, with height \(h\), thickness \(\Delta x\), and approximate radius \(r\). Cutting the shell and laying it flat forms a rectangular solid with length \(2\pi r\), height \(h\) and depth \(dx\). Thus the volume is \(V \approx 2\pi rh\ dx\); see Figure \(\PageIndex{2c}\). (We say "approximately" since our radius was an approximation.)

By breaking the solid into \(n\) cylindrical shells, we can approximate the volume of the solid as

$$V = \sum_{i=1}^n 2\pi r_ih_i\ dx_i,\]

where \(r_i\), \(h_i\) and \(dx_i\) are the radius, height and thickness of the \(i\,^\text{th}\) shell, respectively.

This is a Riemann Sum. Taking a limit as the thickness of the shells approaches 0 leads to a definite integral.

7.3.2.png

Figure \(\PageIndex{2}\): Determining the volume of a thin cylindrical shell.}\label{fig:soupcan}

Key Idea 25: Shell Method

Let a solid be formed by revolving a region \(R\), bounded by \(x=a\) and \(x=b\), around a vertical axis. Let \(r(x)\) represent the distance from the axis of rotation to \(x\) (i.e., the radius of a sample shell) and let \(h(x)\) represent the height of the solid at \(x\) (i.e., the height of the shell). The volume of the solid is

\[V = 2\pi\int_a^b r(x)h(x)\ dx.\]

Special Cases:

  • When the region \(R\) is bounded above by \(y=f(x)\) and below by \(y=g(x)\), then \(h(x) = f(x)-g(x)\).
  • When the axis of rotation is the \(y\)-axis (i.e., \(x=0\)) then \(r(x) = x\).

Let's practice using the Shell Method.

Example \(\PageIndex{1}\): Finding volume using the Shell Method

Find the volume of the solid formed by rotating the region bounded by \(y=0\), \(y=1/(1+x^2)\), \(x=0\) and \(x=1\) about the \(y\)-axis.

This is the region used to introduce the Shell Method in Figure \(\PageIndex{1}\), but is sketched again in Figure \(\PageIndex{3}\) for closer reference. A line is drawn in the region parallel to the axis of rotation representing a shell that will be carved out as the region is rotated about the \(y\)-axis. (This is the differential element.)

7.3.3.png

Figure \(\PageIndex{3}\): Graphing a region in Example \(\PageIndex{1}\).

The distance this line is from the axis of rotation determines \(r(x)\); as the distance from \(x\) to the \(y\)-axis is \(x\), we have \(r(x)=x\). The height of this line determines \(h(x)\); the top of the line is at \(y=1/(1+x^2)\), whereas the bottom of the line is at \(y=0\). Thus \(h(x) = 1/(1+x^2)-0 = 1/(1+x^2)\). The region is bounded from \(x=0\) to \(x=1\), so the volume is

\[V = 2\pi\int_0^1 \dfrac{x}{1+x^2}\ dx.\]

This requires substitution. Let \(u=1+x^2\), so \(du = 2x\ dx\). We also change the bounds: \(u(0) = 1\) and \(u(1) = 2\). Thus we have:

\[\begin{align*} &= \pi\int_1^2 \dfrac{1}{u}\ du \\[4pt] &= \pi\ln u\Big|_1^2\\[4pt] &= \pi\ln 2 - \pi\ln 1\\[4pt] &= \pi\ln 2 \approx 2.178 \ \text{units}^3.\end{align*}\]

Note: in order to find this volume using the Disk Method, two integrals would be needed to account for the regions above and below \(y=1/2\).

With the Shell Method, nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, as demonstrated next.

Example \(\PageIndex{2}\): Finding volume using the Shell Method

Find the volume of the solid formed by rotating the triangular region determined by the points \((0,1)\), \((1,1)\) and \((1,3)\) about the line \(x=3\).

The region is sketched in Figure \(\PageIndex{4a}\) along with the differential element, a line within the region parallel to the axis of rotation. In part (b) of the figure, we see the shell traced out by the differential element, and in part (c) the whole solid is shown.

7.3.4a.png

Figure \(\PageIndex{4}\): Graphing a region in Example \(\PageIndex{2}\)

The height of the differential element is the distance from \(y=1\) to \(y=2x+1\), the line that connects the points \((0,1)\) and \((1,3)\). Thus \(h(x) = 2x+1-1 = 2x\). The radius of the shell formed by the differential element is the distance from \(x\) to \(x=3\); that is, it is \(r(x)=3-x\). The \(x\)-bounds of the region are \(x=0\) to \(x=1\), giving

\[\begin{align*} V &= 2\pi\int_0^1 (3-x)(2x)\ dx \\[4pt] &= 2\pi\int_0^1 \big(6x-2x^2)\ dx \\[4pt] &= 2\pi\left(3x^2-\dfrac23x^3\right)\Big|_0^1\\[4pt] &= \dfrac{14}{3}\pi\approx 14.66 \ \text{units}^3.\end{align*}\]

When revolving a region around a horizontal axis, we must consider the radius and height functions in terms of \(y\), not \(x\).

Example \(\PageIndex{3}\): Finding volume using the Shell Method

Find the volume of the solid formed by rotating the region given in Example \(\PageIndex{2}\) about the \(x\)-axis.

The region is sketched in Figure \(\PageIndex{5a}\) with a sample differential element. In part (b) of the figure the shell formed by the differential element is drawn, and the solid is sketched in (c). (Note that the triangular region looks "short and wide" here, whereas in the previous example the same region looked "tall and narrow." This is because the bounds on the graphs are different.)

The height of the differential element is an \(x\)-distance, between \(x=\dfrac12y-\dfrac12\) and \(x=1\). Thus \(h(y) = 1-(\dfrac12y-\dfrac12) = -\dfrac12y+\dfrac32.\) The radius is the distance from \(y\) to the \(x\)-axis, so \(r(y) =y\). The \(y\) bounds of the region are \(y=1\) and \(y=3\), leading to the integral

\[\begin{align*}V &= 2\pi\int_1^3\left[y\left(-\dfrac12y+\dfrac32\right)\right]\ dy \\[4pt]&= 2\pi\int_1^3\left[-\dfrac12y^2+\dfrac32y\right]\ dy \\[4pt] &= 2\pi\left[-\dfrac16y^3+\dfrac34y^2\right]\Big|_1^3 \\[4pt] &= 2\pi\left[\dfrac94-\dfrac7{12}\right]\\[4pt] &= \dfrac{10}{3}\pi \approx 10.472\ \text{units}^3.\end{align*}\]

7.3.5a.png

Figure \(\PageIndex{5}\): Graphing a region in Example \(\PageIndex{3}\)

At the beginning of this section it was stated that "it is good to have options." The next example finds the volume of a solid rather easily with the Shell Method, but using the Washer Method would be quite a chore.

Example \(\PageIndex{4}\): Finding volume using the Shell Method

Find the volume of the solid formed by revolving the region bounded by \(y= \sin x\) and the \(x\)-axis from \(x=0\) to \(x=\pi\) about the \(y\)-axis.

The region and a differential element, the shell formed by this differential element, and the resulting solid are given in Figure \(\PageIndex{6}\).

7.3.6a.png

Figure \(\PageIndex{6}\): Graphing a region in Example \(\PageIndex{4}\)

The radius of a sample shell is \(r(x) = x\); the height of a sample shell is \(h(x) = \sin x\), each from \(x=0\) to \(x=\pi\). Thus the volume of the solid is

\[V = 2\pi\int_0^{\pi} x\sin x\ dx.\]

This requires Integration By Parts. Set \(u=x\) and \(dv=\sin x\ dx\); we leave it to the reader to fill in the rest. We have:

\[\begin{align*} &= 2\pi\Big[-x\cos x\Big|_0^{\pi} +\int_0^{\pi}\cos x\ dx \Big] \\[4pt] &= 2\pi\Big[\pi + \sin x \Big|_0^{\pi}\ \Big] \\[4pt] &= 2\pi\Big[\pi + 0 \Big] \\[4pt] &= 2\pi^2 \approx 19.74 \ \text{units}^3. \end{align*}\]

Note that in order to use the Washer Method, we would need to solve \(y=\sin x\) for \(x\), requiring the use of the arcsine function. We leave it to the reader to verify that the outside radius function is \(R(y) = \pi-\arcsin y\) and the inside radius function is \(r(y)=\arcsin y\). Thus the volume can be computed as

$$\pi\int_0^1 \Big[ (\pi-\arcsin y)^2-(\arcsin y)^2\Big]\ dy.\]

This integral isn't terrible given that the \(\arcsin^2 y\) terms cancel, but it is more onerous than the integral created by the Shell Method.

We end this section with a table summarizing the usage of the Washer and Shell Methods.

Key Idea 26: Summary of the Washer and Shell Methods

Let a region \(R\) be given with \(x\)-bounds \(x=a\) and \(x=b\) and \(y\)-bounds \(y=c\) and \(y=d\).

\[\begin{align*} & \text{Washer Method} & & \text{Shell Method} \\[4pt] \text{Horizontal Axis} \quad & \pi\int_a^b \big(R(x)^2-r(x)^2\big)\ dx & & 2\pi\int_c^d r(y)h(y)\ dy \\[4pt] \\[4pt] \text{Vertical Axis} \quad & \pi \int_c^d\big(R(y)^2-r(y)^2\big)\ dy & & 2\pi\int_a^b r(x)h(x)\ dx \end{align*}\]

As in the previous section, the real goal of this section is not to be able to compute volumes of certain solids. Rather, it is to be able to solve a problem by first approximating, then using limits to refine the approximation to give the exact value. In this section, we approximate the volume of a solid by cutting it into thin cylindrical shells. By summing up the volumes of each shell, we get an approximation of the volume. By taking a limit as the number of equally spaced shells goes to infinity, our summation can be evaluated as a definite integral, giving the exact value.

We use this same principle again in the next section, where we find the length of curves in the plane.

Contributors and Attributions

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License.  http://www.apexcalculus.com/

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Introduction

Parentheses and brackets are very common in mathematical formulas. You can easily control the size and style of brackets in L a T e X ; this article explains how.

Here's an table of listing some common math braces and parentheses used in L a T e X :

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The size of brackets and parentheses can be manually set, or they can be resized dynamically in your document, as shown in the next example:

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The above example produces the following output:

\[ F = G \left( \frac{m_1 m_2}{r^2} \right) \] Notice that to insert the parentheses or brackets, the \left and \right commands are used. Even if you are using only one bracket, both commands are mandatory. \left and \right can dynamically adjust the size, as shown by the next example:

\[\left[ \frac{ N } { \left( \frac{L}{p} \right) - (m+n) } \right]\]

When writing multi-line equations with the align , align* or aligned environments, the \left and \right commands must be balanced on each line and on the same side of & . Therefore the following code snippet will fail with errors:

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The solution is to use "invisible" brackets to balance things out, i.e. adding a \right. at the end of the first line, and a \left. at the start of the second line after & :

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How do I "X" out a cell from corner to corner?

Look under the Format>Cells>Borders option.

You can create a corner to corner X using interior borders.

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Super Bowl 2024: How to watch SpongeBob SquarePants, Patrick Star on Nickelodeon broadcast for Super Bowl 58

How to watch the super bowl family-friendly broadcast on nickelodeon.

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Super Bowl LVIII is finally here, and it will be a special one. That's because Paramount+ will be providing a family-friendly broadcast of the NFL championship game for the first time. The popular Nickelodeon broadcast of NFL holiday games will be heading to Allegiant Stadium for Super Bowl 58 between the San Francisco 49ers and Kansas City Chiefs . The big game kicked off today and is being broadcast on CBS and Nickelodeon, and streamed on  Paramount+ .

Honoring its 25th anniversary, the cast of SpongeBob SquarePants will highlight Nickelodeon's coverage. SpongeBob (voiced by Tom Kenny) and Patrick Star (voiced by Bill Fagerbakke) will join CBS Sports analyst Nate Burleson and play-by-play announcer Noah Eagle to call the game. Sandy Cheeks (voiced by Carolyn Lawrence) will join the group as a sideline reporter. 

The much-loved SpongeBob SquarePants cartoon has reigned as the most-watched animated series for 21 consecutive years, making this Super Bowl alternate broadcast one to watch. This isn't the first foray into the broadcast spectrum for Fagerbakke, who had a viral moment making his broadcasting debut in the NickMas game in 2022. 

Patrick Star really has a future as a play-by-play announcer. 🤩 📺: #DENvsLAR on @Nickelodeon 📱: Stream on NFL+ pic.twitter.com/65Zxrbz4lT — NFL (@NFL) December 25, 2022

"That was just kind of me responding from all the endless trivia and details we as sports fans immerse ourselves in," Fagerbakke told CBS Sports as he laughed reliving the moment. "That was such a great quote a few years back about Russell Wilson to just let Russ cook, so it just bubbled out. I was enjoying the Rams not being awful (Fagerbakke is a Rams fan).

"We're all about the bubbles in Bikini Bottom, Jeff," Kenny added. 

How to watch Super Bowl LVIII

Date:  Sunday, Feb. 11 |  Time:  6:30 p.m. ET Location:  Allegiant Stadium (Las Vegas) TV:  CBS, Nickelodeon |  Stream:  Paramount+ Follow:   CBS Sports App  

Kenny will join Fagerbakke in the broadcast booth for this one, as SpongeBob joins the Nickelodeon family-friendly broadcast for the first time. SpongeBob himself couldn't contain his level of excitement for the opportunity. 

"It's great! I've done motion capture stuff before, but never attached to the biggest live sporting event on Earth," Kenny said. "It's gonna be a ball, we're just gonna be ad libbing in character. 

"You just gotta be SpongeBob, looking around, seeing what's happening, and cracking dumb jokes about it in real time. That's how I spent my entire 61-year lifespan, just looking around and cracking dub jokes about stuff!

"What could possibly go wrong? Don't answer that!" 

Will SpongeBob do his trademark laugh? This will be harder to pull off with motion capture. 

"I don't know. You just brought up an interesting thing I just thought up," Kenny said. "For me to do the laugh, I have to cheat with my hands (does laugh). Does that mean that SpongeBob's hand goes to SpongeBob's Adam's apple? Does he have one?

"You can't do it when we are on the air," Fagerbakke added. "We won't always be in the frame!" 

"Since it's motion capture, everything my body does -- SpongeBob's body does," Kenny said. "I'm gonna have to pick and choose my moments. I was feeling pretty good about this until you just dashed whatever confidence I had!" 

Just like a live sporting event, there's preparation but not a rehearsal to cover a football game. SpongeBob and Patrick practiced the motion capture, but the game itself will play out similar to Jim Nantz, Tony Romo , and Tracy Wolfson on CBS. 

"When you say rehearsed, that implies that we 'hearsed,' " Kenny joked. "We haven't pre-hearsed, hearsed, or rehearsed! All that real time stuff is what sportscasters do all the time. We're just kind doing what they're doing, just in a stupider way."

What can viewers watching the game expect? There will be "Sweet Victory" performed prior to the game, the popular song from SpongeBob played at the conclusion of "Band Geeks," which is arguably the most popular episode in the show's 25-year history. 

It's also Kenny's favorite episode. 

"That got kind of tied into the Super Bowl," Kenny said. "This time you won't get to look at Adam Levine's six-pack while watching it! Not that I sit around and obsessively watch SpongeBob, but if I'm on an airplane and see a kid watching it. I'm kind of like 'tee hee.'

"Band Geeks is the one." 

In addition to Kenny's trademark SpongeBob laugh, Lawrence said he couldn't wait to bring up the Travis Kelce-Taylor Swift relationship. "I know I'm gonna have to say Taylor's man is hotter than a hickory-smoked sausage!' Lawrence laughed as she referred to Sandy Cheeks' famous line. 

SpongeBob SquarePants has had 277 episodes over 25 seasons since it debuted on July 17, 1999. The show has captivated multiple generations of fans and is one of the most valuable entertainment franchises in the world. 

"Because of SpongeBob, I've been ending up in places I never thought I could be: on Broadway, waving a checkered flag at a NASCAR race, at the Super Bowl. The memories keep on coming," Kenny said. "I'm so excited to do the Super Bowl, but I'm even more excited about when the game is over, everything went okay, and we're kicking back and remembering it -- in a positive way!

"Every time a young adult says to me 'thank you for my childhood,' my heart gets all mushy," Fagerbakke said. "That's a fantastic thing to experience."

No matter what happens on Super Bowl Sunday, SpongeBob, Patrick, and Sandy will have fun. The family-friendly telecast of NFL games has been a hit since its inception in the 2020 season, and there's a reason why. 

"That was a revelation for me last year. This incredible, gladiator, spectator concept. They get to have fun!," Fagerbakke said. "How much fun they had and how much they got to enjoy it. 

"The whole CBS Sports crew, they're so good and they're so excited to do this. Why? Because it's silly. They get to do a different kind of football game." 

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Romcom ending: Taylor Swift and Travis Kelce’s big night at the Super Bowl

The pop star was a focal point during the NFL’s biggest night, getting to enjoy a happy ending with her victorious boyfriend

T aylor Swift, firmly in her WAG era, ended weeks of speculation over her Super Bowl attendance on Sunday night. She successfully managed to jet from a Tokyo arena show to Las Vegas to watch her boyfriend, the Kansas City Chiefs tight end Travis Kelce, play in the Super Bowl.

In typical Swift style, the appearance was a tightly choreographed affair that reportedly involved two private jets (one on standby, just in case) to ensure that she arrived in Sin City in time. When Swift hit Allegiant Stadium, the entourage included Blake Lively, Ice Spice and Kelce’s sister-in-law Kylie.

Kelce, an already popular and decorated football player in his own right, has enjoyed an unprecedented PR boost from dating one of America’s biggest pop stars. He, too, showed up in style ahead of the game, wearing a custom bedazzled Amiri combination of a bouclé shirt and matching shimmering trousers. Kelce also toted a Louis Vuitton bag for good measure. The New York Post estimated that this ensemble cost more than $3,000.

#TaylorSwift ’s boyfriend, Travis Kelce, stepped his fashion game up for #SuperBowl LVIII from #LasVegas . 👏 👏 Via @NFL . pic.twitter.com/jE19Aq0V8M — Perez (@ThePerezHilton) February 11, 2024

It was a look no doubt coordinated to complement Swift’s outfit : $695 crystal jeans from Area, cropped corset and a bomber jacket blazed with the Chiefs’ logo. Lest we forget the reason why she was there, Swift also wore a custom necklace and a football-shaped clutch sporting 87 – Kelce’s jersey number.

Since Swift and Kelce debuted their relationship early in the season, certain (curmudgeonly) NFL fans grumbled over the numerous camera shots of the singer’s private box. But during the Super Bowl , there were few cutaways, maybe surprisingly so for the fans hoping for more. Eras Tour: Part II this was not.

Those watching the game for Swift alone had to hit up social media for an endless stream of content. During a pregame performance by Post Malone, Swift and Lively cuddled each other while swaying to the music.

Later, Swift greeted her fellow pop star Lana Del Rey, leaning down from her elevated perch to hug the star, not unlike the way a princess might greet a peasant. (Lana later joined her in the box .)

Taylor Swift & Lana Del Ray at the #SuperBowlLVIII pic.twitter.com/lWdkLXiIOY — Chicks in the Office (@ChicksInTheOff) February 11, 2024

Another highlight: Swift chugging beers with her longtime friend Nashville stylist Ashley Avignone, as caught on the stadium jumbotron. Well, something has to lift your spirit when your boyfriend’s team is down 10-3 at half-time.

Taylor Swift #SuperBowl pic.twitter.com/ifJZpqwzJ1 — 21 (@21metgala) February 12, 2024

A photo of Swift introducing Ice Spice to Kelce’s brother Jason also made the rounds online, highlighting the improbability of their odd coupling. While Swift may have had all the attention, Ice’s presence was particularly delightful.

Like many viewers watching for everything but the game, she seemed to have no clue what was going on, presumably requiring a full explanation of how football works from Avignone at one point.

“what does grrah mean” pic.twitter.com/CsD7LPuVmZ — tortured poet faith⸆⸉ (@outhewoodsyet) February 11, 2024

Cameras caught Kelce yelling at the Chiefs’ head coach, Andy Reid, in an undeniably uncomfortable moment. But some online commentators bizarrely took it one very weird step further, latching on to the scene as evidence of the player’s anger issues.

“These are very disturbing images from Travis Kelce. Are we sure Taylor is safe around this man?” asked one sports account. “If Travis Kelce is this violent with the cameras on him, what about when Taylor Swift is alone with him?” echoed another X (formerly Twitter) user.

Kelce’s visible frustration during the moment might have been incongruous with his desired reputation as a golden retriever boyfriend . But it’s still quite a stretch to call him an abuser just for getting in a ref’s face. The clip gives ammunition to hard-right conservatives who have repeatedly come after the famous couple for what they perceive as liberal leanings.

Kelce advocated for Covid vaccines and supported Black Lives Matter. While Swift typically stays diplomatic and apolitical, she endorsed Biden in 2020, and many – including a livid Donald Trump – believe another endorsement in 2024 is in the works. Apparently, the lovers couldn’t become the hottest pop culture crossover couple without igniting a few uncomfortable conspiracy theories.

After a relatively boring game – at least for Swift fans expecting more from a relatively sleepy Kansas City performance – things perked up near the end of the third quarter. When the Chiefs earned a touchdown, taking over the scoreboard and winning a lead over the San Francisco 49ers, a jubilant Swift and Co jumped around in the private box.

With Swift, Spice, Lively and Avignone hugging each other, it was a picture-perfect visual summery of Swift’s brand: manufactured empowerment, a celebration of pop feminist girl power.

Taylor and friends celebrating the #CHIEFS TOUCHDOWN!!! 🏈🫶🏆 #SuperBowl2024 pic.twitter.com/vrRjhI7jyr — Taylor Swift Updates (@SwiftNYC) February 12, 2024

The drama of the game continued well into the fourth quarter, when Kansas City caught up with the 49ers to tie the game right toward the end, leading to overtime. Though the Chiefs eventually took the title, it was a close call the entire way through – as photographs of Swift’s box show, the singer chewed on her nails during the closest calls. If the Swift/Kelce pairing is a government psyop, as conservatives want to believe, the CIA sure knows how to stage damn good football.

The hours-long extravaganza ended the way many fans hoped it would – Swift and Kelce, winners, sharing a kiss on the field. (We do not speak of that Viva Las Vegas performance.) A real-life romcom played out on screen.

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How to watch Usher’s Super Bowl halftime show replay

Phil Nickinson

Super Bowl 2024 has come and gone. (The Chiefs won. Again.) That means we have another full year to go before the next one — Super Bowl LIX is scheduled for Sunday, February 9, 2025, in New Orleans — though the hype is sure to never actually stop.

And that rings true for the Super Bowl Halftime Show as well. Whether you thought Usher’s 2024 performance was one of the ages — or that it was nostalgia overshadowed by the likes of H.E.R. and Alicia Keys — it’s worth watching again. And it’s pretty easy to get an official replay of the Super Bowl LVIII Halftime Show.

If you’re using an Apple TV box (that is, the hardware from Apple itself and our reigning pick for the best streaming device you can buy ), just head into the Apple TV app and you’ll see Usher front and center. For now, anyway. That’ll likely change at some point, and you might have to search it out. Note that it doesn’t get the same sort of placement on other platforms as it does in the Apple TV app. It’s officially the Apple Music Halftime Show, and so Apple TV gets some special love. (Same goes for the TV app on Apple devices, though. Go figure.)

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Or, you can hit up the NFL YouTube Channel and check out the halftime show performances there.

Given that this is YouTube we’re talking about, though, we can go one better. We can simply embed all the Usher you could possibly handle. We can make it so you don’t have to leave this very website to see Lil John and Will.i.am. We can have Usher’s gloved hand magically appear beneath these very words, getting you to the full halftime show set that much quicker. Isn’t the future great?

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Phil Nickinson

The 2024 Super Bowl is over. Even if your favorite team won, there's an anticlimactic feeling that is hard to get over. Is that all there is? And what's worse, the weekend is over too, and for a lot of people, the prospect of a long work week is too much to bear.

One cure for these post-Super Bowl blues is Netflix. As the world's most popular streamer, it offers a plethora of movies to access anytime you want. But which ones should you watch? Digital Trends has selected five great Netflix movies to watch after the Super Bowl. Moneyball (2011)

We all must face the sad fact that the 2024 Super Bowl is over. The epic clash between the Kansas City Chiefs and the San Francisco 49ers is in the record books and everyone will continue to speculate what next adventure Taylor Swift and Travis Kelce will go on once the limelight is off them.

What's left to do? How about watching some good movies? Amazon Prime Video has plenty of them, from uproarious comedies to probing sci-fi films, and these three movies stand out as ideal to watch after the Super Bowl. A Fish Called Wanda (1988)

February marks the arrival of new TV shows on Netflix. Love is Blind, Netflix's hit reality dating show, returns on Valentine's Day with a group of men and women hitting the pods. Meanwhile, the long-awaited live-action adaptation of Avatar: The Last Airbender premieres February 22. Both series are bound to find a spot in the weekly top 10 list of the most popular shows on Netflix.

Once you move past the marquee shows, you'll realize there are hundreds of other good programs on Netflix. Many of these series are undervalued despite their dedicated audience and critical acclaim. These three underrated shows should be next in your Netflix queue this February. Our picks include a Canadian mockumentary, a sports docuseries, and a thriller from the Money Heist creator. White Lines (2020)

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Godzilla x Kong: The New Empire

Godzilla x Kong: The New Empire (2024)

Two ancient titans, Godzilla and Kong, clash in an epic battle as humans unravel their intertwined origins and connection to Skull Island's mysteries. Two ancient titans, Godzilla and Kong, clash in an epic battle as humans unravel their intertwined origins and connection to Skull Island's mysteries. Two ancient titans, Godzilla and Kong, clash in an epic battle as humans unravel their intertwined origins and connection to Skull Island's mysteries.

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  • Trivia The change in color for Godzilla's atomic breath going from light blue to a reddish tint, signifying a total boost in power, is identical to the Heisei incarnation having the far more destructive Red Spiral Ray as the sheer strength of his attack is more powerful than his default blue-colored atomic breath.

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